Re: Length of array in hash of arrays

[Grimy]

You're aware that none of these solutions actually print the length of the array, right? Well, they do so in this example, but by pure serendipity. The actual “standard” way to get the length of the array would be $#{$g{k}} or 0+@{$g{k}}

[ambrus]

Yes, I'm aware of that. That's why this is post is in Obfuscated Code rather than Meditations. If you want to get the real solution, the simpler

use 5.012; no warnings; my @t = map { rand } 1..60; my %g = (k => \@t)
+; for my $n (0..1e3) { (my $s = $n) =~ y/0-9/ {}@#$%=^~+/; my $c= qq(
+${s}g{k}}); if (60 == eval $c) { warn $c; } }

prints it together with another fake solution.

[Anonymous Monk]

The simplest way must contain the function 'length':

length(!()x@{(\%x)->{k}})

[gurpreetsingh13]

Like this:

$ perl -e 'my %g = (k => [8, 2, 10, 2, 1, 3],l => [10, 7, 9, 0, 1]); f
+oreach $keyval(sort keys %g){print 0+@{$g{$keyval}};}'

[ambrus]

That prints only the length of the first subarray, the one corresponding to the key "k". Let's do better and use the each keyword to print the length of each subarray.

my %g = (k => [8, 2, 10, 2, 1, 3],l => [10, 7, 9, 0, 1]); 
%gkey = sort keys %g; print 0+@{$g{each %gkey}}, "\n";